Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space. Let $(X_n)_{n\in \mathbb{N}}$ be a sequence of real random variables in $L^1(\mathbb{P})$. Each $X_n$ has a mean $c>0$.

Suppose that $(X_n)$ satisfies the strong law of large numbers, i.e., (I omit a.s. when it's clear) $$ \limsup_n |n^{-1} S_n- c|=0, $$ where $S_n = \sum_{i=1}^n X_i$ .

Is it true that $\liminf_n S_n =\infty$? This is intuitively true because $n^{-1} S_n$ concentrates around $c>0$ for large $n$, but I'm not sure how to show this.

I can show that $\limsup_n S_n = \infty$. If $\limsup S_n=k <\infty$, then $$ \frac{S_{n}}{n}\le\frac{\limsup S_{n}}{n}=\frac{k}{n}\to 0, $$ which is a contradiction.

  • $\begingroup$ The assertion $S_n \leq \limsup S_n$ is not correct. $\endgroup$
    – Andrew
    Aug 15, 2023 at 22:08
  • $\begingroup$ Ah I see. Then I don't know where to start for bot limsup and liminf... $\endgroup$
    – keepfrog
    Aug 15, 2023 at 22:16

1 Answer 1


Hint: The answer is YES. Almost surely $\frac {S_n} n >\frac c 2$ for $n$ sufficiently large. This implies $S_n >\frac {nc} 2$ for $n$ sufficiently large, so $S_n \to \infty$ with probability $1$.

  • $\begingroup$ I understand the implication but I'm not sure how to prove $n^{-1}S_n > c/2$ for $n$ sufficiently large. For example, if $n^{-1}S_n=(-1)^{n}$ and $c=1$, then the assertion is not true. For this particular sequence, it would be impossible though, as $X_n$ would not be in $L^1$. However, in general, how can I use the assumption to show the assertion? I understand that if a.s. convergence is defined with lim instead of limsup, it is trivial. $\endgroup$
    – keepfrog
    Aug 16, 2023 at 21:39
  • 1
    $\begingroup$ If $a_n \geq 0$ and $\lim \sup a_n=0$ then $a_n \to 0$. So $\frac 1 n S_n \to 0$ almost surely. @keepfrog $\endgroup$ Aug 16, 2023 at 23:13
  • $\begingroup$ In your example, the hypothesis is not satisfied. Perhaps, you are confusing $\lim \sup$ with $\lim \inf$ @keepfrog $\endgroup$ Aug 16, 2023 at 23:20
  • $\begingroup$ Ah I see and I was confused with the the example $\endgroup$
    – keepfrog
    Aug 16, 2023 at 23:43

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